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CIE 9231 2026 June Paper 23 Q1

A Level / CIE / FP2

CIE 9231 2026 June Paper 23 Paper · Question 1

题目

Problem

Find the roots of the equation w5+32=0w^5+32=0, giving your answers in the form reiθr\mathrm{e}^{\mathrm{i}\theta}, where r>0r>0 and −π≤θ<π-\pi\le\theta<\pi.

(5)
题目中文翻译

求方程 w5+32=0w^5+32=0 的根,将答案写成 reiθr\mathrm{e}^{\mathrm{i}\theta} 的形式,其中 r>0r>0 且 −π≤θ<π-\pi\le\theta<\pi。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

解法一

思路

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把 −32-32 写成模为 3232、辐角为 π\pi 的数。五次根的模为 22,辐角相差 2π5\frac{2\pi}{5},最后选到题给区间。

答题过程

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Since w5=−32=32eiπw^5=-32=32e^{\mathrm{i}\pi}, the roots have modulus 22 and arguments (π+2kπ)/5(\pi+2k\pi)/5, for integer kk. The argument spacing is 2π/52\pi/5. The five distinct arguments in [−π,π)[-\pi,\pi) give

w1= 2e−iπ,w2=2e−3iπ/5,w3= 2e−iπ/5,w4=2eiπ/5,w5= 2e3iπ/5.\begin{align*} w_1=&\,2e^{-\mathrm{i}\pi},\qquad w_2=2e^{-3\mathrm{i}\pi/5},\\ w_3=&\,2e^{-\mathrm{i}\pi/5},\qquad w_4=2e^{\mathrm{i}\pi/5},\\ w_5=&\,2e^{3\mathrm{i}\pi/5}. \end{align*}